Source: Manhattan Prep
Narcisse and Aristide have numbers of arcade tokens in the ratio 7:3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6;5. What is the least number of tokens that Narcisse could have given to Aristide?
A. 9
B. 17
C. 21
D. 27
E. 53
The OA is B
Narcisse and Aristide have numbers of arcade tokens in the ratio 7:3, respectively...
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Solution:BTGmoderatorLU wrote: ↑Sat Jun 27, 2020 3:06 pmSource: Manhattan Prep
Narcisse and Aristide have numbers of arcade tokens in the ratio 7:3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6;5. What is the least number of tokens that Narcisse could have given to Aristide?
A. 9
B. 17
C. 21
D. 27
E. 53
The OA is B
Let 7n and 3n be the number of tokens Narcisse and Aristide respectively had originally and x be the number of tokens Narcisse gave to Aristide. We can create the equation:
(7n - x) / (3n + x) = 6/5
6(3n + x) = 5(7n - x)
18n + 6x = 35n - 5x
11x = 17n
x = 17n/11
Since 17 is not divisible by 11, we see that the least positive value of n is 11. Therefore, the least positive value of x is 17(11)/11 = 17.
Answer: B
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